//You are given an array prices where prices[i] is the price of a given stock on
// the ith day. 
//
// Find the maximum profit you can achieve. You may complete at most two transac
//tions. 
//
// Note: You may not engage in multiple transactions simultaneously (i.e., you m
//ust sell the stock before you buy again). 
//
// 
// Example 1: 
//
// 
//Input: prices = [3,3,5,0,0,3,1,4]
//Output: 6
//Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 
//3-0 = 3.
//Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
// 
//
// Example 2: 
//
// 
//Input: prices = [1,2,3,4,5]
//Output: 4
//Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 
//5-1 = 4.
//Note that you cannot buy on day 1, buy on day 2 and sell them later, as you ar
//e engaging multiple transactions at the same time. You must sell before buying a
//gain.
// 
//
// Example 3: 
//
// 
//Input: prices = [7,6,4,3,1]
//Output: 0
//Explanation: In this case, no transaction is done, i.e. max profit = 0.
// 
//
// Example 4: 
//
// 
//Input: prices = [1]
//Output: 0
// 
//
// 
// Constraints: 
//
// 
// 1 <= prices.length <= 105 
// 0 <= prices[i] <= 105 
// 
// Related Topics 数组 动态规划 
// 👍 793 👎 0


package leetcode.editor.cn;

//Java：Best Time to Buy and Sell Stock III
 class P123BestTimeToBuyAndSellStockIii {
    public static void main(String[] args) {
        Solution solution = new P123BestTimeToBuyAndSellStockIii().new Solution();
        // TO TEST
    }

    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        public int maxProfit(int[] prices) {
            int n = prices.length;
            int max_k = 2;
            int[][][] dp = new int[n][max_k + 1][2];
            for (int i = 0; i < n; i++)
                for (int k = max_k; k >= 1; k--) {
                    if (i - 1 == -1) {
                        dp[i][k][0] = 0;
                        dp[i][k][1] = -prices[i];
                        continue;
                    }
                    dp[i][k][0] = Math.max(dp[i - 1][k][0], dp[i - 1][k][1] + prices[i]);
                    dp[i][k][1] = Math.max(dp[i - 1][k][1], dp[i - 1][k - 1][0] - prices[i]);
                }
            return dp[n - 1][max_k][0];
        }
        private   int maxProfitII(int[] prices) {
            int hold1 = Integer.MIN_VALUE, hold2 = Integer.MIN_VALUE, release1 = Integer.MIN_VALUE, release2 = Integer.MIN_VALUE;
            if (prices.length == 0) return 0;
            for (int i : prices) {
                hold1 = Math.max(hold1, -i);
                release1 = Math.max(release1, hold1 + i);
                hold2 = Math.max(hold2, release1 - i);
                release2 = Math.max(release2, hold2 + i);
            }
            return release2;
        }
    }
//leetcode submit region end(Prohibit modification and deletion)

}